2.710 Optics
Quiz 1 Solutions
Spring ‘09
1. (a) Without correction, the focus is before the retina, leading to blurred vision:
With correction, the image forms (focused) on the retina:
1
1
1
1
1
1
50 − 45
5
+
=
⇒
=
−
=
=
−fc + 15 50
45
−fc + 15
45 50
50 × 45
50 × 45
50 × 45
−fc + 15 =
= 450 ⇒ fc = −435 (Pc ≈ −2.3D)
5
(b) Draw again the ray-tracing diagram for the object at ∞:
α1 = −
x
x
x
=−
=
fc
−435
435
Draw the imaging system formed by EL above, with object = the virtual image
found by CL.
1
Angular magnification MA = −
450
= −9
50
⇒ α2 = −9 · α1 = −
9x
x
= − 1
435
48 3
From �ICA (see diagram on previous page),
x
= tan α2 = α2 (paraxial approximation)
EFL
x
1
= 1 ⇒ EFL = 48
3
48 3
(c) 1st method: Using the 2nd PP and EFL
1
xi = α · (EFL) = 48 α
3
2nd method: Using the imaging condition
2
x0 = αfc = −435α
The lateral magnification of EL’s imaging system (with the virtual image formed
by CL as the object and real image formed on the retina) is given by:
50
1
1
= − ⇒ xi = ML x0 = − · (−435α)
9
9
450
435
1
=
α = 48 α ⇒ Image is erect.
9
3
ML = −
(d) Use the eye’s pupil as the object and flip the optical system for proper ray-tracing.
1
1
1
15 × 435
+ =
⇒z=−
15 z
−435
450
∴ virtual image created as shown
x�p
z
435
87
= ML = − =
=
xp
15
450
90
The myopic person’s eyes appear smaller and erect.
2. (a) First consider an on-axis point at the object plane.
Clearly, S1 limits the angle of acceptance, so S1 is the Aperture Stop. Now
consider an off-axis point object. The chief ray goes through the center of the
Aperture Stop:
Clearly, S2 limits the angle of acceptance of the chief ray, so S2 is the Field Stop.
3
(b) To find the Entrance Pupil, image S1 (the Aperture Stop) through L1 (flip for
proper use of the sign conventions):
1
1
1
1
1
2
1
+ =
⇒ =
−
= − ⇒ z = −f1
f1 /2 z
f1
f1 f1
f1
z
The virtual image is formed (to the left) as shown.
To find the Exit Pupil, image S1 through S2:
f1
2
�
�
f1
f1
+
f
−
f
�
�
1
1
1
1
1
1
f
2
2
2
�
�
+ =
⇒ = − f1
= 2 f1
= � f12+2f2 � ⇒ z = f2 1 + 2
>0
f2
f2 2 + f2
f1
z
+ f2 z
f2 ( 2 + f2 )
f2
2
The image is real (to the right) as shown.
S2 is the Field Stop and also the Exit Window. S2 is imaged onto the object plane,
where the Entrance Window is located. So the completely annotated system is:
4
(c) From Fig. 1, NA =
considerations:
a1
is the angle of acceptance. For the FoV, there are two
f1
i. Size of Exit Window. S2 is imaged onto the object plane. Flipping for proper
sign conventions, we find the lateral magnification as |a�2 | = ff12 a2 :
ii. Angle of Chief Ray. For this we reconsider the object plane and the Entrance
Pupil:
a�
1
We can see that FoV = 2 =
2f1
2f1
�
f1
a2
f2
�
=
1 a2
.
2 f2
(d) Even though S2 is the Field Stop, the location of S1 has the unfortunate conse­
quence of limiting the FoV because of the bending of the chief ray as we saw in
the previous question.
We can see easily that if we move S1 to the right, the CR to the left of L1 becomes
closer to horizontal and the FoV increases. Its maximum value will be a2 /f2 when
S1 is located symmetrically f1 to the right of L1. (The EnP, ExP are then at ∞.
See the solutions for the 2.71 exam for more details.) Moving the Aperture Stop
(i.e. S1) further to the right again reduces the FoV.
The symmetric placement of the AS (f1 to the right of L1, f2 to the left of L2)
also has some desirable aberration reduction effects, e.g. it reduces distortion.
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Spring 2009
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